Find sum of count of duplicate numbers in all subarrays of given array

Last Updated : 15 Feb, 2024

Given an array arr[] of size N. The task it to find the sum of count of duplicate numbers in all subarrays of given array arr[]. For example array {1, 2, 3, 2, 3, 2} has two duplicate elements (i.e, 2 and 3 come more than one time in the array).

Examples:

Input: N = 2, arr = {3,3}
Output: 1
Explanation: All subarrays of the arr are {3}, {3,3}, {3} have values 0, 1, 0 respectively, so the answer is 1.

Input: N = 4, arr = {1, 2, 1, 2}
Output: 4
Explanation: All subarrays of the arr are {1},{1,2},{1,2,1},{1,2,1,2},{2},{2,1},{2,1,2},{1},{1,2},{2} have values 0,0,1,2,0,0,1,0,0,0 respectively, so answer is 4.

Approach: This can be solved with the following idea:

To avoid considering all possible subarrays, we can use below approach:

We can first create a map that stores the positions of each distinct element in the array. Then, for each distinct element, we calculate the contribution of subarrays that contain this element. This avoids the need to iterate through all subarrays explicitly.

Step-by-step approach:

Initialize an unordered map mp of arrays m to store all positions of each distinct element in the array.
Iterate through the array arr and populate the map. For each element, append its position to the corresponding entry in the map.
Initialize a variable ans for answer.
Iterate through the entries in the mp:
- If an element appears only once, continue to the next element as there can be no subarray with distinct elements that appear more than once.
- For elements that appear more than once, calculate their contribution to the count of subarrays:
  - Initialize a variable prev to -1 to represent the position before the first occurrence of the element.
  - Iterate through the positions of the element and calculate the contribution of subarrays containing this element. The contribution is determined by the product of the distance to the previous occurrence of the element and the distance to the next occurrence.
  - Update ans with the calculated contribution, taking the modulo operation to avoid integer overflow.
  - Update prev to the current position.
Return the final value of ans.

Below is the implementation of the above approach:

C++

// C++ Implementation #include <bits/stdc++.h> #include <iostream> using namespace std;  // Function to find the values of subarrays int ValueOfSubarrays(int N, vector<int>& arr) {      int i = 0;      // Initailze a map     unordered_map<int, vector<int> > mp;      // Iterate in array     while (i < arr.size()) {          mp[arr[i]].push_back(i);         i++;     }      i = 0;     int mod = 1e9 + 7;     long long prev = -1;     long long ans = 0;      // Interate in map     for (auto a : mp) {         prev = -1;          // I frequency is 1         if (a.second.size() == 1) {             continue;         }         int j = 0;          // Iterate to find the impact of each element         while (j < a.second.size() - 1) {             ans += (a.second[j] - prev)                    * (N - a.second[j + 1]);             ans %= mod;             prev = a.second[j];             j++;         }     }      // Return the value     return ans; }  // Driver code int main() {      int N = 5;     vector<int> arr = { 1, 2, 2, 3, 2 };      // Function call     cout << ValueOfSubarrays(N, arr);     return 0; }

Java

import java.util.*;  public class SubarraysValue {      // Function to find the values of subarrays     static int valueOfSubarrays(int N, ArrayList<Integer> arr) {          int i = 0;          // Initialize a map         HashMap<Integer, ArrayList<Integer>> mp = new HashMap<>();          // Iterate in array         while (i < arr.size()) {             if (!mp.containsKey(arr.get(i))) {                 mp.put(arr.get(i), new ArrayList<>());             }             mp.get(arr.get(i)).add(i);             i++;         }          i = 0;         int mod = (int) 1e9 + 7;         long prev = -1;         long ans = 0;          // Iterate in map         for (Map.Entry<Integer, ArrayList<Integer>> entry : mp.entrySet()) {             prev = -1;              // If frequency is 1             if (entry.getValue().size() == 1) {                 continue;             }             int j = 0;              // Iterate to find the impact of each element             while (j < entry.getValue().size() - 1) {                 ans += (entry.getValue().get(j) - prev) * (N - entry.getValue().get(j + 1));                 ans %= mod;                 prev = entry.getValue().get(j);                 j++;             }         }          // Return the value         return (int) ans;     }      // Driver code     public static void main(String[] args) {          int N = 5;         ArrayList<Integer> arr = new ArrayList<>(Arrays.asList(1, 2, 2, 3, 2));          // Function call         System.out.println(valueOfSubarrays(N, arr));     } }

Python3

def value_of_subarrays(N, arr):     # Initialize a dictionary to store indices of each element     mp = {}      # Iterate in array     i = 0     while i < len(arr):         if arr[i] not in mp:             mp[arr[i]] = []         mp[arr[i]].append(i)         i += 1      i = 0     mod = 10**9 + 7     prev = -1     ans = 0      # Iterate in the dictionary     for key, value in mp.items():         prev = -1          # If frequency is 1, no impact on subarrays         if len(value) == 1:             continue          j = 0          # Iterate to find the impact of each element         while j < len(value) - 1:             ans += (value[j] - prev) * (N - value[j + 1])             ans %= mod             prev = value[j]             j += 1      # Return the value     return ans  # Driver code if __name__ == "__main__":     N = 5     arr = [1, 2, 2, 3, 2]      # Function call     print(value_of_subarrays(N, arr))

using System; using System.Collections.Generic;  public class SubarraysValue {     // Function to find the values of subarrays     static int ValueOfSubarrays(int N, List<int> arr)     {         int i = 0;          // Initialize a dictionary         Dictionary<int, List<int>> mp = new Dictionary<int, List<int>>();          // Iterate in array         while (i < arr.Count)         {             if (!mp.ContainsKey(arr[i]))             {                 mp[arr[i]] = new List<int>();             }              mp[arr[i]].Add(i);             i++;         }          i = 0;         int mod = (int)1e9 + 7;         long prev = -1;         long ans = 0;          // Iterate in dictionary         foreach (var a in mp)         {             prev = -1;              // If frequency is 1             if (a.Value.Count == 1)             {                 continue;             }              int j = 0;              // Iterate to find the impact of each element             while (j < a.Value.Count - 1)             {                 ans += (a.Value[j] - prev) * (N - a.Value[j + 1]);                 ans %= mod;                 prev = a.Value[j];                 j++;             }         }          // Return the value         return (int)ans;     }      // Driver code     public static void Main()     {         int N = 5;         List<int> arr = new List<int> { 1, 2, 2, 3, 2 };          // Function call         Console.WriteLine(ValueOfSubarrays(N, arr));     } } //this code is contributed by Aman

JavaScript

function valueOfSubarrays(N, arr) {     // Initialize a Map to store indices of each element     let mp = new Map();      // Iterate in array     for (let i = 0; i < arr.length; i++) {         if (!mp.has(arr[i])) {             mp.set(arr[i], []);         }         mp.get(arr[i]).push(i);     }      let mod = 10**9 + 7;     let prev = -1;     let ans = 0;      // Iterate in the Map     for (let [key, value] of mp) {         prev = -1;          // If frequency is 1, no impact on subarrays         if (value.length === 1) {             continue;         }          let j = 0;          // Iterate to find the impact of each element         while (j < value.length - 1) {             ans += (value[j] - prev) * (N - value[j + 1]);             ans %= mod;             prev = value[j];             j++;         }     }      // Return the value     return ans; }  // Driver code let N = 5; let arr = [1, 2, 2, 3, 2];  // Function call console.log(valueOfSubarrays(N, arr));

Output

Time Complexity: O(N)
Auxiliary Space: O(N)

Find sum of count of duplicate numbers in all subarrays of given array

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Find sum of count of duplicate numbers in all subarrays of given array

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